Question

A wet bar of soap slides down a ramp 9.0 m long inclined at 8.0°. How long does it take to reach the bottom? Assume${\mu }_k=\mathbf{0}\mathbf{.}\mathbf{060}$.

Short answer

The time taken by the soap bar to reach the bottom of the ramp is 4.80 s.

Step-by-step solution

Step 1. Understanding the forces acting on the soap

A wet soap bar slides down the ramp inclined at an8.0°angle.Since the soap slides downward on the ramp, a kinetic frictional force acts on the soap in the direction opposite to the bar’s motion.

Resolve the mass components along the x-axis and y-axisdirections. Then, apply the equilibrium conditions and evaluate the time taken by the soap to reach the bottom of the ramp.

Step 2. Identification of given data 

The given data can be listed below as:

• The length of the ramp is $L=9\mathrm{m}$.
• The value of the kinetic friction coefficient is ${\mu }_k=0.060$.
• The angle of inclination of the ramp is $\theta =8°$.
• The acceleration due to gravity is $g=9.81\mathrm{m}/{\mathrm{s}}^2$.
• The initial velocity of the bar is $u=0\mathrm{m}/\mathrm{s}$.

Step 3. Free body diagram representation of the wet soap bar on the ramp

The free body diagram of the soap bar can be shown as:

Here, $F_N$ is the normal reaction force on the soap bar, $F_{fr}$ is the frictional force, g is the acceleration due to gravity, mgis the weight of the bar, and m is the mass of the bar.

Step 4. Determination of the acceleration of the wet soap bar 

The bar is sliding down the ramp with some acceleration (a).

At the equilibrium condition, the forces along the horizontal direction can be expressed as:

$\begin{array}{c}\sum F_x=ma\\ mg\sin \theta -F_{fr}=ma\\ F_{fr}=mg\sin \theta -ma\dots \left(i\right)\end{array}$

At the equilibrium condition, the forces along the vertical direction can be expressed as:

$\begin{array}{c}\sum F_y=0\\ mg\cos \theta -F_N=0\\ F_N=mg\cos \theta \dots \left(ii\right)\end{array}$

The frictional force can be expressed as:

$F_{fr}={\mu }_k{F}_N$

Substitute the values of (i) and (ii) in the above expression.

$\begin{array}{c}mg\sin \theta -ma={\mu }_{k}mg\cos \theta \\ a=g\sin \theta -{\mu }_{k}g\cos \theta \end{array}$

Substitute the values in the above expression.

$\begin{array}{c}a=9.81\mathrm{m}/{\mathrm{s}}^2\times \sin 8°-0.060\times 9.81\mathrm{m}/{\mathrm{s}}^2\times \cos 8°\\ =0.78\mathrm{m}/{\mathrm{s}}^2\end{array}$

Step 5. Determination of the time taken by the wet soap bar to reach the bottom of the ramp

The distance moved by the soap can be expressed as:

$L=ut+\frac{1}{2}a{t}^2$

Here, L is the distance moved by the wet soap bar, which is equal to the length of the bar,t is the time taken by the bar to reach the bottom, and $u$is the initial velocity of the bar whose value is 0 m/s.

Substitute the value of the initial velocity of the bar in the above expression.

$\begin{array}{c}L=0\mathrm{m}/{\mathrm{s}}^2\times t+\frac{1}{2}a{t}^2\\ \frac{1}{2}a{t}^2=L-0\\ t^2=\frac{2L}{a}\\ t=\sqrt{\frac{2L}{a}}\end{array}$

Substitute the value of the initial velocity of the bar in the above expression.

$\begin{array}{c}t=\sqrt{\frac{2\times 9\mathrm{m}}{0.78\mathrm{m}/{\mathrm{s}}^2}}\\ =4.80\mathrm{s}\end{array}$

Thus, the time taken by the soap bar to reach the bottom is 4.80 s.